Question

Question 6 (i)
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar(DOC) = ar(AOB)

Answer 1

Let us draw $DN\perp ACandBM\perp AC$.
(i) In $\Delta DON$ and $\Delta BOM$,
$\angle DNO=\angle BMO={90}^{\circ }$ (By construction)
$\angle DON=\angle BOM$ (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule,$\Delta DON$ $\cong \Delta BOM$
$\Rightarrow DN=BM...\left(1\right)$
We know that congruent triangles have equal areas.
Area $\left(\Delta DON\right)=Area\left(\Delta BOM\right)...\left(2\right)$
In $\Delta DNC$ and $\Delta BMA$,
$\angle DNC=\angle BMA$ (By construction)
CD = AB (Given)
DN = BM [Using equation (1)]
$\Delta DNC\cong \Delta BMA$ (RHS congruence rule)
$\Rightarrow Area\left(\Delta DNC\right)=Area\left(\Delta BMA\right)...\left(3\right)$
On adding equations (2) and (3), we obtain,
Area $\left(\Delta DON\right)+Area\left(\Delta DNC\right)=Area\left(\Delta BOM\right)+Area\left(\Delta BMA\right)$
Therefore, Area $\left(\Delta DOC\right)=Area\left(\Delta AOB\right)$